simpson's rule (1 Viewer)

[red802]

hey guys, is it just easier to use the complex simpson rule than the easier one because it works for large and small numbers

 

[SoulSearcher]

It really depends on the person.

I use the long version myself, because I think its easier.

Others may use the shorter version because they think its easier.

It depends

 

[Mountain.Dew]

use the shorter method. in any HSC level maths exam, they wont give u many intervals or ranges of values to do, that is painful. perhaps only 2 or 3 x-values is enough.

therefore, i advise u follow this equation:

Area ~ = 1/6(b-a)[f(a) + f(b) + 4([b+a]/2)], where b and a is the x-values of the interval.

 

[Dreamerish*~]

I always used the long method, because usually the question has many intervals.

If you use the short one, you'd have to keep repeating.

 

[pLuvia]

I use this one

b
∫ f(x) dx ≈ h/3 [f(a) + 4[f(x₁) + f(x₃) + f(x₅)...] + 2[f(x₂) + f(x₄) + f(x₆)....] + f(b)]
a

Where h=(b-a)/2

With my one in textbooks they usually expand the whole equation out making it really difficult to remember, but this way you can see a pattern in it. Hope that helped

 

[m0ofin]

What exactly are the long and short methods? I always use Area ~ = h/3 [(First + Last) + 2(sum of odd ordinates) + 4(sum of even ordinates)] where h = (b-a)/n Which is pretty much the same as Pluvia's if you take y₀ as y₁.

 

[Riviet]

I use pLuvia's method, easy way to remember it is:

Area=h/3.[first + last + 4x(odds) + 2x(evens)], where first and last refer to the first and last function values, odds and evens refer to the odd numbered functions and even numbered functions.

N.B the odd and even function values exclude the first and last function values, so for example: if 5,6,7,8,9 are your function values where 5 is the first and 9 is your last, 6 is your first odd function, 7 is your first even function, and 8 is your second odd function.
I'll clarify my point with an illustrration:

5|6|7|8|9|
F|O|E|O|L|
F|1|2|3|L| -> notice how 1 is odd, 2 is even 3 is odd.

5,6,7,8,9 are your function values, F is the first, L is last, O is odd, E is even.

 

[pLuvia]

What exactly are the long and short methods? I always use Area ~ = h/3 [(First + Last) + 2(sum of odd ordinates) + 4(sum of even ordinates)] where h = (b-a)/n Which is pretty much the same as Pluvia's

You got that wrong moofin

Area ~ = h/3 [(First + Last) + 2(sum of even ordinates) + 4(sum of odd ordinates)] where h = (b-a)/n

It should be this, careless mistakes

 

[pLuvia]

What exactly are the long and short methods? I always use Area ~ = h/3 [(First + Last) + 2(sum of odd ordinates) + 4(sum of even ordinates)] where h = (b-a)/n Which is pretty much the same as Pluvia's

You got that wrong moofin

Area ~ = h/3 [(First + Last) + 2(sum of even ordinates) + 4(sum of odd ordinates)] where h = (b-a)/n

It should be this, careless mistakes

 

[m0ofin]

No, I'm sure it works out to be the same as long as you take the first interval to be y1 instead of y0.

For example, if the intervals were 1, 2, 3, 4, 5 and 6:

(y₀, y₁, y₂, y₃, y₄, y₅) Using h/3[ F + L + 2(even) + 4 (odd)]
= h/3 [y₀ + y₅ + 2(y₂ + y₄) + 4(y₁ + y₃)]
=h/3 [1 + 6 + 2(3 + 5) + 4(2 + 4)]

(y₁, y₂, y₃, y₄, y₅, y₆) Using h/3[ F + L + 2(odd) + 4(even)]
= h/3 [y₁ + y₆ + 2(y₃ + y₅) + 4(y₂ + y₄)]
= h/3 [1 + 6 + 2(3 + 5) + 4(2 + 4)]

 

[Riviet]

No, I'm sure it works out to be the same as long as you take the first interval to be y1 instead of y0.

For example, if the intervals were 1, 2, 3, 4, 5 and 6:

(y₀, y₁, y₂, y₃, y₄, y₅) Using h/3[ F + L + 2(even) + 4 (odd)]
= h/3 [y₀ + y₅ + 2(y₂ + y₄) + 4(y₁ + y₃)]
=h/3 [1 + 6 + 2(3 + 5) + 4(2 + 4)]

You haven't substituted the numbers in correctly, your last line should read:

h/3 [1 + 6 + 2(2 + 4) + 4(3 + 5)]
=/= h/3 [1 + 6 + 2(3 + 5) + 4(2 + 4)]

Hence it doesn't work both ways, it has to be 4xodds and 2xevens.

 

[m0ofin]

You haven't substituted the numbers in correctly, your last line should read:

h/3 [1 + 6 + 2(2 + 4) + 4(3 + 5)]
=/= h/3 [1 + 6 + 2(3 + 5) + 4(2 + 4)]

Hence it doesn't work both ways, it has to be 4xodds and 2xevens.

No, it was substituted correctly and both ways work, depending on whether you use y₀ or y₁ first. Clicky.
1 | 2 | 3 | 4 | 5 | 6
y₀ | y₁ | y₂ | y₃ | y₄ | y₅ - 2 even and 4 odd
y₁ | y₂ | y₃ | y₄ | y₅ | y₆ - 2 odd and 4 even

 

[pLuvia]

What do you mean if you use y₀ or y₁ first, if you are using the formula

Area ~ = h/3 [(First + Last) + 2(sum of even ordinates) + 4(sum of odd ordinates)] where h = (b-a)/n

As long as you know the first number is y₀ or something equivalent, the second number, y₁ should go with the 4(sum of odd ordinates) and the next, y₂, should go with the 2(sum of even ordinates) then it should be fine

BUT y₀ is always the first number in the equation, why would you change it to y₁ that just contradicts the y₀ since that one is the initial, and y₁ is the number after it

 

[webby234]

I find it easier to use the table and just do them in groups of three

eg
x 1 2 3 4 5
y 1 4 9 16 25
...1 4 1
.........1 4 1

so then multiply by h/3

= 1/3 ((1 + 4*4 + 9) + (9 + 4*16 + 25))


100 posts

 

[Riviet]

Alright fair enough with the 4 even and 2 odd version, but there's no point in knowing both, sooner or later, one day, you will get confused and get it wrong. So stick to the formula you personally prefer, and learn that one.

 

[SoulSearcher]

Alright fair enough with the 4 even and 2 odd version, but there's no point in knowing both, sooner or later, one day, you will get confused and get it wrong. So stick to the formula you personally prefer, and learn that one.

Agreed. And thats why I'll continue to use the long version

 

[m0ofin]

Alright fair enough with the 4 even and 2 odd version, but there's no point in knowing both, sooner or later, one day, you will get confused and get it wrong. So stick to the formula you personally prefer, and learn that one.

It does help to skim through both to see which one works for you but of course, it is much easier to just stick to what you knew first. Thanks for the tip though